A
Number Trick

Select
a three digit number in which the first and the last digit differ by at
least two. Construct a second number by reversing the order of digits
in the first. Form a third number by taking the difference of the first
two. Reverse the order of digits in the third number to construct a fourth,
and add the third and fourth number. The result is the number 1089.

Example:
Select the number 732. Note: 7 - 2 = 5 > 2, making this number a valid
choice. Next, construct the number 237 by reversing the order of digits.
Take the difference:

732
- 237 = 495

From
495, construct the number 594, and add:

495
+ 594 = 1089.

Proof:
Let the originally selected number be represented by

100a
+ 10b + c

1.

where
a, b, c, are integers between 0 and 9 inclusive, and occupy the hundreds,
tens, and units places, respectively. We may assume, without loss of generality,
that a > c. Now, construct a new number by reversing the order of digits

100c
+ 10b + a

2.

Take
the difference

100(a
- c) + (c - a)

3.

and simplify
to

99(a
- c) = 99

4.

where
= a - c. We would like to represent this number, algebraically, as a number
in base ten; i.e., with a form similar to that shown in 1.
We begin by rewriting 99
as

99
= (10 . 9)
+ (1 . 9)

5.

Since
2
9 by hypothesis, we know that 18
9
81 so that we may write

9
= 10
+ v

6.

where
and v are integers between 0 and 9 inclusive, and occupy the tens and
units places, respectively. Thus, the Right Hand Side of eq.
5 may be rewritten as

(10
. 9)
+ (1 . 9)
= [10 . (10
+ v)] + [1 . (10
+ v)]

=
100
+ 10(
+ v) + v

7.

If the
integer
+ v does not exceed a single digit, then the number,

100
+ 10(
+ v) + v

is the
number sought. Let us find all possible values of
+ v for 2
9:

9

+ v

2

18

9

3

27

9

4

36

9

5

45

9

6

54

9

7

63

9

8

72

9

9

81

9

Thus,
+ v does not exceed 9 (in fact, it equals 9 in all cases). We conclude
that the number, 100
+ 10(
+ v) + v, is the one we sought, and correctly represents the difference,
99,
as a number in base ten, with
occupying the hundreds place; (
+ v), the tens place; and v, the units place. The number obtained by reversing
the order of its digits is then

100v
+ 10(
+ v) +

8.

and the
sum of these last two numbers is

100(
+ v) + 20(
+ v) + (
+ v) = 121(
+ v)

9.

But,
for all values of a considered, (
+ v)= 9, so that 121(
+ v) = 1089.